Page 65 - Op Amps Design, Applications, and Troubleshooting

48 AMPLIFIERS

Let us compute the actual load voltage in Figure 2.3 at a frequency of 5 kilohertz.
First we compute the ideal output voltage Equation (2,1):

We have found the value of r o at 5 kilohertz to be 4.96 ohms. Using the method
shown in Figure 2.4, we can now determine the actual load voltage with Equation
(2.17).

At a frequency of 5 kilohertz when the output resistance has increased to nearly
5 ohms, the effect of nonideal output resistance is minimal. Problems could be
anticipated when the output resistance exceeds 1 percent of the value of load
resistance.
Although the preceding calculation illustrates the effects of output resis-
tance, it is valid only if we are below the frequency that causes slew rate limiting
(/SRI)- If/ssL is exceeded, we can expect the actual output to be much lower than the
value computed with Equation (2.17), and the output will be nonsinusoidal in
shape. Additionally, this method is inappropriate if the output drive capability of
the op amp is exceeded.

Output Current Capability. The output of the op amp in Figure 2.3 must sup-
ply two currents: the current through the feedback resistor (i F) and the current to
the load resistor (z' L). It is the sum of these currents that flows into or out of the out-
put of the op amp.
The output of many (but not all) op amps is short circuit protected. That is,
the output may be shorted directly to ground or to either DC supply voltage with-
out damaging the op amp. For a protected op amp (such as the 741), the output
current capability is not determined by the maximum allowable current before
damage, but rather depends on the amount of reduced output voltage the appli-
cation can tolerate.
With no output current being supplied to the load, the output voltage stays
at the expected v 0 level, and the total output current is equal to i F. As the load cur-